Hdoj 2602 Bone Collector "Dynamic Planning"

Topic Links:

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=17434

Dynamic planning:

Classic 01 Knapsack problem

There are n items and a backpack with a capacity of V. The cost of article I is c[i], the value is w[i].
The sum of the values is maximized by solving which items are loaded into the backpack.

This is the most basic knapsack problem, characterized by: only one piece of each item, you can choose to put or not put.
Define the state with sub-question: F[i][v] Indicates the maximum value that the first I item can get in a backpack with a capacity of V. The state transfer equation is:
F[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}
Explain it in more detail:
"Put the first I item into a backpack with a capacity of V" This sub-problem, if only consider the article I item strategy (put or not), then can be converted into a only 1 items in the first I-related issues.
If I do not put the article I, then the problem is converted to "the first I-1 items into the capacity of the backpack", the value of f[i-1][v];
If you put the item I, then the problem is converted to "the first I-1 items into the remaining capacity of v-c[i] in the backpack", the maximum value can be obtained at this time is f[i-1][v-c[i]], plus the value obtained by placing the article I item w[i].

First consider how to achieve the basic idea above, there must be a main loop I=1..N, each time the two-dimensional array f[i][0..v] all values.
So, if you only use an array of f[0..v], can you guarantee that the f[v] in the end of the first cycle is the state f[i][v] that we define?
F[I][V] (F[i-1][v] and f[i-1] [V-c[i]] Two sub-problems are recursive, can be guaranteed in the push F[i][v] (also in the main cycle of the first time) can get f[v] and f[i-1][v] [f[i-1]] value?
In fact, this requires us to v=v in each main loop. 0 in the order of the push f[v], so as to ensure that push f[v] f[v-c[i]] saved is the value of state F[i-1][v-c[i]].
The pseudo code is as follows:
For I=1..N
For V=v. 0
F[v]=max{f[v],f[v-c[i]]+w[i]};
One of the F[v]=max{f[v],f[v-c[i]]} is just equivalent to our transfer equation F[i][v]=max{f[i-1][v],f[i-1][v-c[i]]}, because now F[v-c[i]] is equivalent to the original f[i-1][v-c [i]].

The last f[v] is the optimal solution to the problem.

1#include <cstdio>2#include <cstring>3 Const intMAXL = ++5;4 intMaxintAintb) {returnA>b?a:b;}5 intT, N, V, Value[maxl], Volume[maxl], DP[MAXL];6 7 intMain () {8scanf"%d", &t);9      while(t--){Tenscanf"%d%d", &n, &v); One          for(inti =1; I <= N; i++) scanf ("%d", &value[i]); A          for(inti =1; I <= N; i++) scanf ("%d", &volume[i]); -Memset (DP,0,sizeof(DP)); -          for(inti =1; I <= N; i++){ the              for(intj = V; J >= Volume[i]; j--) {//Note that the stop condition of J, one is to prevent cross-border, two is less than 0 after the answer is DP[J], has been calculated -                 //The inverse of the Traverse from V to 0 guarantees that Dp[j-volume[i]] is the result of the previous layer, -                 //and because the second dimension of the DP array involves only the previous layer of data, the data from the other layers is no longer accessible -                 //thus eliminating the first dimension of the DP array, each time the previous data is refreshed, saving space complexity to O (n); +DP[J] = max (Dp[j], dp[j-volume[i]]+value[i]); -             } +         } Aprintf"%d\n", Dp[v]); at     } -  -     return 0; -}

　　

Hdoj 2602 Bone Collector "Dynamic Planning"